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📐 Mathematics  ·  Conic Sections  ·  JEE

Find the equation of the hyperbola with foci (±5, 0) and transverse axis of length 8.

Answer: x 2 /16 - y 2 /9 = 1.

  • A x<sup>2</sup>/16 - y<sup>2</sup>/9 = 1
  • B x<sup>2</sup>/9 - y<sup>2</sup>/16 = 1
  • C x<sup>2</sup>/16 + y<sup>2</sup>/9 = 1
  • D x<sup>2</sup>/25 - y<sup>2</sup>/9 = 1

Correct answer: A. x<sup>2</sup>/16 - y<sup>2</sup>/9 = 1

Explanation: Foci (±5,0) means c=5. Transverse axis length=2a=8, so a=4, a²=16. b²=c²−a²=25−16=9. Standard form (foci on x-axis): x²/16−y²/9=1. Ans: x²/16−y²/9=1.

The Four Conics by EccentricityCircle (e=0)Ellipse (0<e<1)Parabola (e=1)Hyperbola (e>1)

All four conics form a single family distinguished only by eccentricity: a circle is the most "closed" (e=0), an ellipse is an elongated closed curve, a parabola is the borderline open curve (e=1), and a hyperbola has two separate open branches (e>1).

Concept context

Study circles, parabolas, ellipses, and hyperbolas as curves formed by intersecting a plane with a double cone, with their standard equations and key properties.

Read the full Conic Sections notes →